Advertisements
Advertisements
प्रश्न
If sinA + sin2A = 1, then the value of the expression (cos2A + cos4A) is ______.
पर्याय
1
`1/2`
2
3
Advertisements
उत्तर
If sinA + sin2A = 1, then the value of the expression (cos2A + cos4A) is 1.
Explanation:
Given,
sinA + sin2A = 1
⇒ sinA = 1 – sin2A = cos2A ...[∵ sin2θ + cos2θ = 1]
On squaring both sides, we get
sin2A = cos4A
⇒ 1 – cos2A = cos4A
⇒ cos2A + cos4A = 1
APPEARS IN
संबंधित प्रश्न
If `sec alpha=2/sqrt3` , then find the value of `(1-cosecalpha)/(1+cosecalpha)` where α is in IV quadrant.
Show that `sqrt((1+cosA)/(1-cosA)) = cosec A + cot A`
Prove the following trigonometric identity.
`cos^2 A + 1/(1 + cot^2 A) = 1`
Prove the following identities:
`(cosecA)/(cosecA - 1) + (cosecA)/(cosecA + 1) = 2sec^2A`
If x = r sin A cos B, y = r sin A sin B and z = r cos A, then prove that : x2 + y2 + z2 = r2
Prove that:
`1/(sinA - cosA) - 1/(sinA + cosA) = (2cosA)/(2sin^2A - 1)`
`(sec^2 theta-1) cot ^2 theta=1`
`(sectheta- tan theta)/(sec theta + tan theta) = ( cos ^2 theta)/( (1+ sin theta)^2)`
Simplify : 2 sin30 + 3 tan45.
If a cos θ + b sin θ = 4 and a sin θ − b sin θ = 3, then a2 + b2 =
Prove the following identity :
`(cosecA - sinA)(secA - cosA) = 1/(tanA + cotA)`
Prove the following identity :
`sqrt((1 - cosA)/(1 + cosA)) = sinA/(1 + cosA)`
If sec θ = `25/7`, then find the value of tan θ.
Prove the following identities.
`(1 - tan^2theta)/(cot^2 theta - 1)` = tan2 θ
a cot θ + b cosec θ = p and b cot θ + a cosec θ = q then p2 – q2 is equal to
Choose the correct alternative:
Which is not correct formula?
tan2θ – sin2θ = tan2θ × sin2θ. For proof of this complete the activity given below.
Activity:
L.H.S = `square`
= `square (1 - (sin^2theta)/(tan^2theta))`
= `tan^2theta (1 - square/((sin^2theta)/(cos^2theta)))`
= `tan^2theta (1 - (sin^2theta)/1 xx (cos^2theta)/square)`
= `tan^2theta (1 - square)`
= `tan^2theta xx square` .....[1 – cos2θ = sin2θ]
= R.H.S
If tan θ = 3, then `(4 sin theta - cos theta)/(4 sin theta + cos theta)` is equal to ______.
Show that: `tan "A"/(1 + tan^2 "A")^2 + cot "A"/(1 + cot^2 "A")^2 = sin"A" xx cos"A"`
Show that, cotθ + tanθ = cosecθ × secθ
Solution :
L.H.S. = cotθ + tanθ
= `cosθ/sinθ + sinθ/cosθ`
= `(square + square)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ............... `square`
= `1/sinθ xx 1/square`
= cosecθ × secθ
L.H.S. = R.H.S
∴ cotθ + tanθ = cosecθ × secθ
